The problem is as follows:
A particle of mass $m$ is tied to a very thin wire. Assume the wire is inflexible and of negligible mass. The particle is spinning about a fixed axis as shown located in the center of the circle. Let $T_a$ and $T_b$ be the modulus of the tensions in the string when the particle is located in the points $a$ and $b$ respectively. Find the mass $m$ of the particle if the difference between the tensions $T_{b}-T_{a}=39.2\,N$. Assume $g=9.8\frac{m}{s^2}$
The alternatives are as follows:
$\begin{array}{ll} 1.&0.8\,kg\\ 2.&2.0\,kg\\ 3.&5.8\,kg\\ 4.&1.0\,kg\\ 5.&0.5\,kg\\ \end{array}$
For this problem what I've attempted to do was to use the force at the top to be as follows:
$T+mg=\frac{mv_a^2}{R}$
At this point the Tension must be zero (I don't know if this statement is correct.)
This reduces the top equation to:
$v_a=\sqrt{Rg}$
Then to obtain the speed in the lowest point would be by the conservation of mechanical energy:
$E_i=E_f$
$\frac{1}{2}mv^2_{a}+mgR=mg(-R)+\frac{1}{2}mv_{b}^2$
Then inserting in the above equation would give the speed for the bottom:
Cancelling masses and multiplying by $2$ to both terms:
$v^2_{a}+4gR=v^2_{b}$
Since it is known $v_{a}$ then:
$v_{b}^2=Rg+4Rg=5Rg$
Finally I'll use these in the given statements:
The tension in the top:
$T_a+mg=\frac{mv_{a}^2}{R}$
Tension in the bottom.
$T_b-mg=\frac{mv_{b}^2}{R}$
Doing a difference between these:
$T_b-T_a=2mg+\frac{mv_{b}^2}{R}-\frac{mv_{a}^2}{R}$
Replacing the known values:
$T_b-T_a=2mg+\frac{m(5Rg)}{R}-\frac{m(Rg)}{R}$
$T_b-T_a=2mg+4mg=6mg$
Then:
$39.2=6m(10)$
Which results into:
$m=0.67\,kg$
But the answers sheet indicates that the mass is $2\,kg$.
For doing that what it should happenned is that the "$2mg$" is negative in the right side of the equation but I can't find a way to do that. Can someone help me here?. Is it me?, or did I overlooked anything?. Help.
Applying Newton's second law in (a) and (b) we have:
$$ T_a + m\,g = m\,\frac{v_a^2}{R}\,, \; \; \; \; \; \; T_b - m\,g = m\,\frac{v_b^2}{R} $$
and subtracting the first equation from the second, we have:
$$ T_b - T_a = 2\,m\,g + m\,\frac{v_b^2 - v_a^2}{R}\,. $$
So, imposing the conservation of mechanical energy between (a) and (b), we have:
$$ \frac{1}{2}\,m\,v_a^2 + m\,g\,R = \frac{1}{2}\,m\,v_b^2 - m\,g\,R $$
from which:
$$ v_b^2 - v_a^2 = 4\,g\,R\,. $$
In conclusion, we get:
$$ T_b - T_a = 6\,m\,g \; \; \; \Rightarrow \; \; \; (39.2\,N) = 6\,m\,(9.8\,m/s^2) \; \; \; \Leftrightarrow \; \; \; m \approx 0.67\,kg\,, $$
as you correctly calculated!